Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Giải:
\(\dfrac{1}{3}-\dfrac{1}{12}-\dfrac{1}{20}-...-\dfrac{1}{90}-\dfrac{1}{110}=x-\dfrac{5}{13}\)
\(\Leftrightarrow\dfrac{1}{1.3}-\dfrac{1}{3.4}-\dfrac{1}{4.5}-...-\dfrac{1}{9.10}-\dfrac{1}{10.11}=x-\dfrac{5}{13}\)
\(\Leftrightarrow\dfrac{-1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}\right)=x-\dfrac{5}{13}\)
\(\Leftrightarrow\dfrac{-1}{2}\left(\dfrac{1}{1}-\dfrac{1}{11}\right)=x-\dfrac{5}{13}\)
\(\Leftrightarrow\dfrac{-1}{2}.\dfrac{10}{11}=x-\dfrac{5}{13}\)
\(\Leftrightarrow\dfrac{-5}{11}=x-\dfrac{5}{13}\)
\(\Leftrightarrow\dfrac{-5}{11}+\dfrac{5}{13}=x\)
\(\Leftrightarrow x=\dfrac{-10}{143}\)
Vậy ...
\(\frac{1}{3}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}-\frac{1}{72}-\frac{1}{90}-\frac{1}{110}=x-\frac{5}{13}\)
\(\frac{1}{3}-\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\right)=x-\frac{5}{13}\)
\(\frac{1}{3}-\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\right)=x-\frac{5}{13}\)
\(\frac{1}{3}-\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\right)=x-\frac{5}{13}\)
\(\frac{1}{3}-\left(\frac{1}{3}-\frac{1}{11}\right)=x-\frac{5}{13}\)
\(\frac{1}{3}-\frac{1}{3}+\frac{1}{11}=x-\frac{5}{13}\)
\(\frac{1}{11}=x-\frac{5}{13}\)
\(x=\frac{1}{11}+\frac{5}{13}\)
\(x=\frac{68}{143}\)
Lời giải:
a, Ta có: \(A=\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+...+\frac{1}{22}>\frac{1}{22}+\frac{1}{22}+\frac{1}{22}+\frac{1}{22}+...+\frac{1}{22}=\frac{1}{22}.11=\frac{11}{22}=\frac{1}{2}\)
Vậy: \(A>\frac{1}{2}\)
b, Ta có: \(B=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{99}+\frac{1}{100}\)
\(=\left(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{49}+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\)
Mà: \(\left(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{49}+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\text{}\text{}\text{}>\left(\frac{1}{50}+...+\frac{1}{50}+\frac{1}{50}\right)+\left(\frac{1}{100}+...+\frac{1}{100}+\frac{1}{100}\right)\)
=> \(B\text{}\text{}\text{}>\frac{1}{50}.41+\frac{1}{100}.50=\frac{41+25}{50}=\frac{33}{25}>1\)
Vậy: \(B>1\)
c, Ta có: \(C=\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{16}+\frac{1}{17}< \frac{1}{5}+\frac{1}{6}+\left(\frac{1}{7}+...+\frac{1}{7}+\frac{1}{7}\right)=\frac{11}{30}+11.\frac{1}{7}=\frac{407}{210}< \frac{420}{210}=2\)
Vậy: \(C< 2\)
Chúc bạn học tốt!
Tick cho mình nhé!
A=1
A >-19A=-19
\(A=1\)
\(A>-19A=-19\)
K NHA